# Show that Y is a finite set

• Mar 31st 2010, 03:02 PM
tn11631
Show that Y is a finite set
Let X and Y be sets satisfying Y $\subset$X $\subset$(- $\infty$, $\infty$). Suppose that X is compact and that for every x $\in$X, there is a neighborhood of x which contains only a finite number of points of Y. This number may vary with x. Show that Y is a finite set.

What I have so far is that since X is compact it is closed and bounded and since it is closed there is an accumulation point in each neighborhood of X. then the neighborhood contains the point (x- $\epsilon$,x+ $\epsilon$. And the Ix's are open sets that contain X.

I'm not sure what to do next and/or how to actually do the proof of it, hope someone can help. thanks
• Mar 31st 2010, 03:46 PM
Drexel28
Quote:

Originally Posted by tn11631
Let X and Y be sets satisfying Y $\subset$X $\subset$(- $\infty$, $\infty$). Suppose that X is compact and that for every x $\in$X, there is a neighborhood of x which contains only a finite number of points of Y. This number may vary with x. Show that Y is a finite set.

What I have so far is that since X is compact it is closed and bounded and since it is closed there is an accumulation point in each neighborhood of X. then the neighborhood contains the point (x- $\epsilon$,x+ $\epsilon$. And the Ix's are open sets that contain X.

I'm not sure what to do next and/or how to actually do the proof of it, hope someone can help. thanks

Let $N_x$ denote the neighborhood of $x$ whose intersection with $Y$ is finite. Clearly, $\left\{N_x\right\}_{x\in Y}$ is an open cover for $Y$ and so by assumption it has a finite subcover $N_{x_1},\cdots,N_{x_n}$. It follows that $\text{card }Y=\text{card }Y\cap\bigcup_{j=1}^{n}N_{x_j}=\text{card }\bigcup_{j=1}^{n}\left(N_{x_j}\cap Y\right)\leqslant\sum_{j=1}^{n}\text{card }\left(N_{x_j}\cap Y\right)<\infty$
• Mar 31st 2010, 05:04 PM
shuxue
what does "card" mean? thanks
• Mar 31st 2010, 05:06 PM
Tinyboss
Cardinality--the number of elements in a set.
• Mar 31st 2010, 05:27 PM
Drexel28
Alternatively, Bolzano-Weierstrass property since we are discussing metric spaces.
• Mar 31st 2010, 05:28 PM
Tinyboss
I liked your first one much better. Not only more general but easier.
• Mar 31st 2010, 05:34 PM
Drexel28
Quote:

Originally Posted by Tinyboss
I liked your first one much better. Not only more general but easier.

Oh well. I made a typo in it also. Let the OP figure it out.
• Mar 31st 2010, 07:37 PM
tn11631
Quote:

Originally Posted by Drexel28
Alternatively, Bolzano-Weierstrass property since we are discussing metric spaces.

for the Bolzano-Weierstrass would we be using the contrapositive of the theorem? Also, is there a different way to do this proof besides the way you showed? just curious. Thanks
• Mar 31st 2010, 07:47 PM
Tinyboss
B-W says that every bounded sequence in R^n has a convergent subsequence. So suppose Y is infinite. Then Y has a countably infinite subset. Enumerate it as $\{y_n\}_{n=1}^\infty$. Then there's a convergent subsequence $y_{n_k}$. But since X is closed and every element of the sequence is in X, the limit is in X. Let L be the limit. Then every neighborhood of L contains infinitely many elements of the sequence, which is a contradiction.
• Mar 31st 2010, 07:48 PM
Drexel28
Quote:

Originally Posted by Tinyboss
B-W says that every bounded sequence in R^n has a convergent subsequence. So suppose Y is infinite. Then Y has a countably infinite subset. Enumerate it as $\{y_n\}_{n=1}^\infty$. Then there's a convergent subsequence $y_{n_k}$. But since X is closed and every element of the sequence is in X, the limit is in X. Let L be the limit. Then every neighborhood of L contains infinitely many elements of the sequence, which is a contradiction.

That's not actually the B.W.P. I was thinking. The one I was thinking is that every infinite subset of a compact metric space has a limit point. (though the proof is the same)
• Mar 31st 2010, 07:53 PM
tn11631
Quote:

Originally Posted by Tinyboss
B-W says that every bounded sequence in R^n has a convergent subsequence. So suppose Y is infinite. Then Y has a countably infinite subset. Enumerate it as $\{y_n\}_{n=1}^\infty$. Then there's a convergent subsequence $y_{n_k}$. But since X is closed and every element of the sequence is in X, the limit is in X. Let L be the limit. Then every neighborhood of L contains infinitely many elements of the sequence, which is a contradiction.

Thats what i meant, contradiction! lol So if i were to show this with B-W-T do I still need to state that since X is closed there is an accumulation point in each neighborhood of X and that the neighborhood $\subset$(x-epsilon, x+epsilon)? or can am I allowed to just go straight into the B-W?
• Mar 31st 2010, 07:58 PM
Tinyboss
Rather than say there's an accumulation point in some neighborhood, you want to first show that (at least) one accumulation point exists in X (L in my post above), then say that any neighborhood of this point must contain infinitely many elements of Y, contrary to the hypothesis. You don't need any epsilonics here. (Or maybe you do...if this is all analysis and no topology...in that case, yeah, say that $(L-\varepsilon,L+\varepsilon)\cap X$ contains a point of Y distinct from L, for any epsilon, so in fact it contains infinitely many for any epsilon.)
• Mar 31st 2010, 08:06 PM
tn11631
Quote:

Originally Posted by Tinyboss
Rather than say there's an accumulation point in some neighborhood, you want to first show that (at least) one accumulation point exists in X (L in my post above), then say that any neighborhood of this point must contain infinitely many elements of Y, contrary to the hypothesis. You don't need any epsilonics here. (Or maybe you do...if this is all analysis and no topology...in that case, yeah, say that $(L-\varepsilon,L+\varepsilon)\cap X$ contains a point of Y distinct from L, for any epsilon, so in fact it contains infinitely many for any epsilon.)

Thanks, its all coming together now. I was thinking about B-W, I actually have it written down in the scribbles of my notes for this problem lol Thanks for clearing this problem up for me!